Motor and ESC Theory - Current Draw
Motor and ESC Theory - Current Draw
(OP)
Hi guys,
I'm trying to understand (DC brushed) motor behaviour in certain situations when coupled with a PWM ESC (powered by 24V battery). The following situation is not realistic but will help me to understand, I'd appreciate any replies.
Imagine my motor is in an application (a wheeled vehicle) where drag force of the vehicle is constant, and unrelated to vehicle speed. Again, not necessarily true in the real world, but play along :)
Let's say, we are trundling along at 40mph, 100% PW, with the full 24V from my battery. Drag force (at all speeds, remember) is 40N, giving 40A current draw from the motor.
Now let's say I halve the duty cycle of the PWM, 50%, giving only an effective 12V to motor. This in turn decreases speed (20mph). Drag force though, stays the same, at 40N, so the motor, technically, still wants to draw 40A. What current does the motor actually get? Does the PWM affect the current the motor receives? If so, how?
Thanks in advance,
Chris
I'm trying to understand (DC brushed) motor behaviour in certain situations when coupled with a PWM ESC (powered by 24V battery). The following situation is not realistic but will help me to understand, I'd appreciate any replies.
Imagine my motor is in an application (a wheeled vehicle) where drag force of the vehicle is constant, and unrelated to vehicle speed. Again, not necessarily true in the real world, but play along :)
Let's say, we are trundling along at 40mph, 100% PW, with the full 24V from my battery. Drag force (at all speeds, remember) is 40N, giving 40A current draw from the motor.
Now let's say I halve the duty cycle of the PWM, 50%, giving only an effective 12V to motor. This in turn decreases speed (20mph). Drag force though, stays the same, at 40N, so the motor, technically, still wants to draw 40A. What current does the motor actually get? Does the PWM affect the current the motor receives? If so, how?
Thanks in advance,
Chris





RE: Motor and ESC Theory - Current Draw
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Motor and ESC Theory - Current Draw
- current and torque are proportional
- voltage and speed are proportional
RE: Motor and ESC Theory - Current Draw
I know that the ESC will change the effective voltage from the batteries. This is how motor speed is changed, because "voltage and speed are proportional". So on a 50% PWM duty cycle, the battery's 24V is "averaged out" to 12V as far as the motor is concerned, hence speed is halved.
I am wondering if the ESC does the same thing to the current flow. Surely the current supply also varies according to the PWM duty cycles. So with this in mind, I can picture two possible outcomes to my hypothetical situation. These I will list below, after copying in my original questions as a refresher:
"Now let's say I halve the duty cycle of the PWM, 50%, giving only an effective 12V to motor. This in turn decreases speed (20mph). Drag force though, stays the same, at 40N, so the motor, technically, still wants to draw 40A. What current does the motor actually get?"
Outcome 1 - Current to the motor stays at an average of 40A, due to PWM peaks of 80A (troughs of 0).
Outcome 2 - PWM peaks of 40A are seen, but due to the 50% duty cycle (and troughs of 0), an average current draw of only 20A is seen at the motor.
Can anyone tell me which outcome will occur? Or will another outcome that I have not foreseen occur?
Thanks,
Chris
RE: Motor and ESC Theory - Current Draw
RE: Motor and ESC Theory - Current Draw
What about current flow through the speed controller then. When we plot voltage versus time through the ESC, you see some square-shaped peaks and troughs, right? What would a similar plot for current look like?
RE: Motor and ESC Theory - Current Draw
RE: Motor and ESC Theory - Current Draw
Imagine your motor is hooked up to a dyno which has the ability to over drive the motor.
The motor is generating almost as much voltage as is applied to it. When the motor is running on the dyno at no load the current will be small. If the motor is then driven faster, the back EMF (the voltage generated by the motor) will rise and the current will drop.
You will soon reach a speed where the current is zero. If the speed is increased further the motor becomes a generator and has the ability to drive current through the source. (Whether it does depends on the source.
Now start to load the motor. As the load increases the speed drops and the back EMF drops. When the back EMF drops relative to the applied voltage, the current increases. With 12 Volts applied the difference between 12 volts and the back EMF is almost the same as the difference between 24 Volts and the back EMF at the same torque load.
The relationship between applied torque load and speed drop is almost constant across a very broad speed range. It is the speed difference between no load and full load which will determine the current.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Motor and ESC Theory - Current Draw
Vm = Ke*w + I*R
The motor torque equation for generated torque T is:
T = Kt*I
In your first case, you are supplying 48 volts to the motor. Some of this is to "overcome" the back EMF of the motor when the vehicle is moving at 40mph (Ke*w). The rest is to push current through the winding (I*R).
Now you reduce the supply voltage to 24 volts. It will no longer be able to push current through the winding at this speed, so less torque will be generated. Since the load torque is constant, the car will decelerate, reducing the back EMF, until the back EMF is low enough that you have enough voltage headroom to push enough current through the motor to generate torque matching the load torque. This speed will be LESS than half of the steady state speed from 48 volts.
To throw some semi-plausible numbers at this scenario:
40 mph = 17.78 m/sec
Using an effective drive radius of 0.01m (through gearing), the motor is turning at 283 rad/sec (2700 rpm) at this speed.
Using a back EMF constant Ke of 0.15V/(rad/sec), the back EMF at this speed is 42.45V.
Kt must be 0.15 N*m/A. To produce a force F of 40N with a drive radius of 0.01m, the torque must be 0.04 N*m, and the current must be 0.04/0.15 = 0.266A.
At 48V supply with a back EMF of 42.45V, the remaining voltage is 5.55V to produce this 0.266A, giving a resistance of 20.8 ohms (oh well...)
At a supply voltage of 24V, when producing enough torque to match the 40N load force, there is still an I*R drop of 5.55V. So the back EMF will be 24-5.55=18.45V, and the motor speed will be 18.45/0.15 = 123 rad/sec. This corresponds to a vehicle speed of 17.4mph, a little less than half of the 40mph at twice the supply voltage.
(Feel free to check my math for errors. I did this quickly.)
Curt Wilson
Delta Tau Data Systems
RE: Motor and ESC Theory - Current Draw
Imagine your motor is hooked up to a dyno which has the ability to over drive the motor.
The motor is generating almost as much voltage as is applied to it. When the motor is running on the dyno at no load the current will be small. If the motor is then driven faster, the back EMF (the voltage generated by the motor) will rise and the current will drop.
You will soon reach a speed where the current is zero. If the speed is increased further the motor becomes a generator and has the ability to drive current through the source. (Whether it does depends on the source.
Now start to load the motor. As the load increases the speed drops and the back EMF drops. When the back EMF drops relative to the applied voltage, the current increases. With 12 Volts applied the difference between 12 volts and the back EMF is almost the same as the difference between 24 Volts and the back EMF at the same torque load.
The relationship between applied torque load and speed drop is almost constant across a very broad speed range. It is the speed difference between no load and full load which will determine the current.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Motor and ESC Theory - Current Draw
Explaining how/why a H bridge or PWM circuit can reduce the voltage while the current stays the same is beyond the scope of an "Engineering Tips" web site.
RE: Motor and ESC Theory - Current Draw
The first part of my question was "what happens". I didn't even know that the current would stay constant, even when a PWM is involved (I thought that might change things w.r.t the DC motor rules). "How does it happen" was the second part of my question. I'm sorry, and with all respect, but your reply didn't help clearly answer either. Please don't be offended by me saying that.
To everyone else, thanks for descriptions and links. I'm still reading through most of them. I'll respond (probably with more questions) once I'm up to speed!
RE: Motor and ESC Theory - Current Draw
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Motor and ESC Theory - Current Draw
RE: Motor and ESC Theory - Current Draw
RE: Motor and ESC Theory - Current Draw
RE: Motor and ESC Theory - Current Draw
RE: Motor and ESC Theory - Current Draw
Once you understand the motor terminal voltage equation and play what ifs with it enough times, the pwm width question should be a no brainer and you probably will ask yourself why you even asked it to begin with, since it has nothing to do with your motor understanding.
Good luck and happy learning!
www.KilroyWasHere<dot>com
RE: Motor and ESC Theory - Current Draw
Curt's equations for shunt dc motor:
Vm = Ke*w + I*Ra
T = Kt*I
Solve each of above equations for I, and equate the two resulting expressions:
T/Kt = (Vm-Ke*w)/Ra
Solve for w:
w = (Vm - Ra* T/Kt)/Ke
This equation could represent a speed-vs-torque curve for constant voltage Vm, or a speed-vs-Vm curve for constant torque T. We're looking at the constant T case for op.
What is the behavior of the equation w = (Vm - Ra* T/Kt)/Ke ?
For Ra* T/Kt<<Vm, then w~Vm. This is the approximation that Lionel mentioned. Drawing w vs Vm would be straight line through the origin with positive constant slope.
OTOH, considering the effects of non-neglible Ra* T/Kt, thhe relationship w vs Vm would be another line a fixed distance below the first line drawn above. Since that fixed distance is a higher percentage at low values, if you double (from 12 to 24 volts), you more than double the result.
Under our assumption of constant torque, then what happens I?
I = T/Kt.
Nothing happens to I (it depends on torque, not affected by speed if torque held constant). That's the other thing Lionel said. And this one is exact (to within the assumptions of the model).
Summary of the two things Lionel mentioned that we proved:
I ~T is exact (within the model)
w~V is approximate (assumes Ra* T/Kt<<Vm)
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(2B)+(2B)' ?
RE: Motor and ESC Theory - Current Draw
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(2B)+(2B)' ?
RE: Motor and ESC Theory - Current Draw
T = Kt*Ia
could also be rewritten:
T = K *Phi * Ia
If permanent magnet motor, then Phi is a constant.
If shunt field, then Phi depends of field current which changes with the varying voltage (unless we provide separate voltage inputs to armature and field).
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(2B)+(2B)' ?
RE: Motor and ESC Theory - Current Draw
Ke=82.3*Kt for a 3ph AC synchronous sine driven motor when Ke is Vrms/Krpm & Kt is my beloved #-ft/a
Ke=0.79*Kt for a 3ph AC synchronous sine driven motor when Ke is Vrms/rad/sec & Kt is nm/a
Ke=Kt for a 3ph AC synchronous 6 step driven motor when Ke is Vrms/rad/sec & Kt is nm/a
Ke=Kt for a DC PM motor when Ke is Vrms/rad/sec & Kt is nm/a
Although there are even more other variations of the constant for other variations of units (Vpeak vs Vrms, etc), both terms are still the same.
www.KilroyWasHere<dot>com
RE: Motor and ESC Theory - Current Draw
So instead of T = Kt*Ia, I'd write T = Kt' * Phi * Ia (where Kt' = Kt/Phi)
and instead of Eg = Ke*w, I'd write Eg = Ke' * Phi * w (where Ke' = Ke/Phi)
The rough equivalence of the constants involved in torque equation and the voltage equation can be seen as follows:
T = Ke' * Phi * Ia
Eg = Kt' * Phi * w
Multiply the first equation by w and the 2nd equation by Ia
w*T = w*(Ke'*Phi * Ia)
Ia*Eg =Ia*( Kt'*Phi* w)
Now the LHS of first expression (w*T) is output power while the LHS of 2nd expression (Ia*Eg) is input power. Neglecting losses, we could equate those LHS and equate the RHS:
w*(Ke'* Phi * Ia) = Ia*( Kt'* Phi * w)
Cancel out all the common terms:
Ke' = Kt'
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(2B)+(2B)' ?
RE: Motor and ESC Theory - Current Draw
I don't know anything about dc motors in automotive applications. Can anyone shed some light on that?
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(2B)+(2B)' ?
RE: Motor and ESC Theory - Current Draw
0.79 = pi/4 ... correct? Why was I thinking the relationship was Ke=0.866*Kt?
RE: Motor and ESC Theory - Current Draw
http://www.baldor.com/downloads/manuals/_downloads......
matches up with some of my Kollmorgen info and it has been a few years since I derived it....
www.KilroyWasHere<dot>com
RE: Motor and ESC Theory - Current Draw
(I don't know how to 'quote' someone)
Do you mean what kind are u sed in automotive apps like electric seats and windows?
Used to be series wound field type (at least in my old 1964 chevy and chrysler), but all new ones I have seen are PM magnet (recall GM says sued original inventor and patentor Kollmorgen saying they invented Neodimium Iron Boron magnets with their magna-quench process) so Curt's equation is 100% accurate.
www.KilroyWasHere<dot>com
RE: Motor and ESC Theory - Current Draw
Equation 23 on page 20 was the relationship that I was recalling.
RE: Motor and ESC Theory - Current Draw
If you're interested, try this:
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That will show up as follows (you can preview to check):
You can get more help with "TGML" (Tecumseh Group Markup Language... because this is a Tecumseh Group site) by clicking the question-mark to the left of "preview" in your posting screen.
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(2B)+(2B)' ?
RE: Motor and ESC Theory - Current Draw
Thank you electricpete.
www.KilroyWasHere<dot>com
RE: Motor and ESC Theory - Current Draw
The equations to use in analysis here are independent of the method of modulation to control the supply voltage -- it does not matter if pulse-width modulation, linear modulation, or some other method is used. Any transient variation introduced by PWM is of far too short a time period to have any effect on what we are talking about here.
My equations did assume a permanent magnet field for the brushed motor. Some of the same effects would be in play for other types of brushed motors, but the details would be different.
One of the key ideas to come out of the analysis is that the motor velocity is only truly proportional to the supply voltage in the no-load case.
As Mike Kilroy pointed out, Ke and Kt are really the same quantity, as they must be for conservation of energy. In consistent units, such as SI for DC motors, they will have the same numerical values. As several have pointed out, things get really strange for brushless motors. The Baldor reference is a good one; here is another good one just written by a Kollmorgen app engineer:
http://machinedesign.com/controllers/servo-paramet...